The transformed equation of 3x^2 - 4xy = r^2 | vedclass

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Question

(C) Let the angle of rotation be $	heta = \operatorname{Tan}^{-1}(2)$. Then $	an 	heta = 2$. Using the trigonometric identities,$\sin 	heta = \fra

Vedclass · Accepted Answer

$4y^2 - x^2 = r^2$

Vedclass · Answer

(C) Let the angle of rotation be $	heta = \operatorname{Tan}^{-1}(2)$. Then $	an 	heta = 2$. Using the trigonometric identities,$\sin 	heta = \frac{2}{\sqrt{5}}$ and $\cos 	heta = \frac{1}{\sqrt{5}}$. The transformation equations for rotation of axes are $x = X \cos 	heta - Y \sin 	heta$ and $y = X \sin 	heta + Y \cos 	heta$. Substituting the values,$x = \frac{X - 2Y}{\sqrt{5}}$ and $y = \frac{2X + Y}{\sqrt{5}}$. Substitute these into the given equation $3x^2 - 4xy = r^2$: $3 \left( \frac{X - 2Y}{\sqrt{5}} ight)^2 - 4 \left( \frac{X - 2Y}{\sqrt{5}} ight) \left( \frac{2X + Y}{\sqrt{5}} ight) = r^2$. $\frac{3}{5} (X^2 - 4XY + 4Y^2) - \frac{4}{5} (2X^2 + XY - 2Y^2) = r^2$. $\frac{1}{5} (3X^2 - 12XY + 12Y^2 - 8X^2 - 4XY + 8Y^2) = r^2$. $\frac{1}{5} (-5X^2 - 16XY + 20Y^2) = r^2$. Wait,re-evaluating the substitution: $3x^2 - 4xy = 3(\frac{X-2Y}{\sqrt{5}})^2 - 4(\frac{X-2Y}{\sqrt{5}})(\frac{2X+Y}{\sqrt{5}}) = \frac{3(X^2-4XY+4Y^2) - 4(2X^2-3XY-2Y^2)}{5} = \frac{3X^2-12XY+12Y^2-8X^2+12XY+8Y^2}{5} = \frac{-5X^2+20Y^2}{5} = 4Y^2 - X^2$. Thus,the transformed equation is $4Y^2 - X^2 = r^2$.

The transformed equation of $3x^2 - 4xy = r^2$ when the coordinate axes are rotated about the origin through an angle of $\operatorname{Tan}^{-1}(2)$ in the positive direction is:

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